I could not however locate a visualization of the wavefunction of a proton. The reason I was looking for one is to see whether the three quarks that make it up "occupy" disjoint regions of space (i.e. the maximum probability locations are separated).

Can someone please point to one such visualization, or explain why they can't be created (or why my question is nonsensical, which is always a possibility...)

Although not a specialist in particle physics, I have a strong feeling that this is, indeed, not possible, because I'd treat quarks in a quantum field approach i.e. in 2nd quantization. Don't really know how to visualize those, as, for example, particle number isn't something that's fixed
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LagerbaerJan 13 '11 at 18:50

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@Lagerbaer: I don't think just 2nd quantization will help you here if by that you mean just perturbative approach to particles. Low energy QCD is quite non-perturbative because of the very strong coupling. If there is a way to answer this question it must be through lattice-QCD or AdS/QCD and it won't be easy. I am looking forward to answers.
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MarekJan 13 '11 at 18:58

Like Marek, I strongly suspect that the only way you could perhaps get a meaningful answer out of this is through lattice QCD. That's not really my specialty so I can't give that answer, but later on perhaps I can try to at least post an explanation of how the proton is described in QCD, which might help clarify why this is such a difficult question.
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David Z♦Jan 13 '11 at 19:09

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@romkyns: yes and no. Quarks are the same as electrons in that they are both elementary particles described by few numbers (mass, spin, charges). But they differ in important way: quarks carry a strong charge and interact by strong force. This force is (sorry for playing Captain obvious) strong at low interaction energies (which are the energies the quarks have inside the proton) and hard to investigate by usual means; in particular, the simple quantum mechanical picture that is used to analyze an atom. You need the full quantum field theory here.
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MarekJan 13 '11 at 22:57

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The problem is that the force between the quarks is so strong that the binding energy is itself sufficiently high to create new particles. This is the reason why, for example, single quarks are never observed: When pulling them apart, the binding energy is converted to new quarks.
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LagerbaerJan 14 '11 at 4:12

3 Answers
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for the electron in the Hydrogen atom, the orbital motion doesn't interact with the electron's spin, so "the wave function" pretty much means just a complex $\psi(x,y,z)$. You may choose the electron's spin to be up or down independently of that.

However, you must realize that a wave function of three quarks has many more components. First of all, there are three particles in relative motion rather than two. It means that even if you decouple the center-of-mass coordinates, the relative wave function is a function of six coordinates, $x_1,y_1,z_1,x_2,y_2,z_2$.

So you can't really visualize the full wave function in a simple way because it is a function of six variables rather than three and of course, it doesn't factorize into a product of functions of three variables (something that is strictly speaking true even for all atoms more complicated than the Hydrogen atom).

Moreover, you must understand that the quarks have many discrete degrees of freedom that are not decoupled from the orbital motion in this case. Each of the three quarks has 1 of 3 colors and 1 of 2 possible values of the spin. The combinations of the color really force you to consider 27 combinations of the quark colors - 27 different wave functions (only 6 of them are nonzero and equal to each other, up to a sign) - and 8 combinations of the three quarks' spin. The spins are correlated as well.

So the right wave function for 3 quarks is really a set of $27 \times 8$ complex functions of six variables. Of course, you may visualize various aspects of these functions by integrating it in space and so on.

It's important that the strong force between the non-relativistic quarks is spin-dependent. That's why the protons' spin is not arbitrary - it equals $1/2$ instead. So one of the quarks' spin differs from the remaining two.

Even if you managed to invent ways how to visualize the $27\times 8$ wave functions of 6 variables describing the relative positions of the three quarks, with the best correlations between the colors, spins, and motion, it would still be a hugely oversimplified model of the proton. In reality, it is not true that the proton only contains 3 quarks. Those 3 quarks we normally talk about are analogous to "valence electrons" in an atom but there is also a large see of equally real gluons and quark-antiquark pairs - additional partons whose total color vanishes.

To really describe the wave function of the proton, you need to talk about the right theory of the proton's structure - quantum chromodynamics (QCD) - which is an example of a quantum field theory - one discovered in the early 1970s. It has infinitely many degrees of freedom, like any field theory, so the right "wave function" is really a "wave functional" or a function of infinitely many variables.

You must realize that many approximations valid in atomic physics break down. For example, the speed of electrons in the atoms is very small - effectively controlled by the fine-structure constant $1/137$ that tells you the speed in the units of the speed of light. That allows you to use non-relativistic quantum mechanics. However, the proton's "strong" fine-structure constant relevant for the "three quarks only" is close to one, so the speed of quarks in the proton is always comparable to the speed of light. Consequently, the extra relativistic energy/mass is comparable to the rest mass as well as the interaction energy, and the proton always has enough energy to create quark-antiquark pairs and so on. It's a mess where relativity is needed much like particle creation and annihilation.

In some sense, it's true that "the 3 quarks", if you select them from the sea of the infinitely many partons, like to occupy three different regions in the ground state. But to be more accurate about what this statement actually means for the wave function(al) of the actual proton, you would have to go through a series of so many approximations and idealizations that it's not worth it. The goal of quantum field theory is not to visualize the structure of something; the goal is to calculate the results of the experiments and one can't really design good experiments that would probe the detailed shape of the wave function of the proton.

The parton distribution functions are the closest observables to this information. However, their very assumption is that the proton has infinitely many rather than 3 partons (quarks, antiquarks, or gluons). To summarize, don't expect anything that is as valid yet as useful as the pictures for the atomic physics. Proton's analogy to an atom has severe limitations.

I have seen this image everywhere, including popular science journals like SciAm. Talk about misleading... They might as well have published a picture of three hamsters running after each other!
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romkynsJan 14 '11 at 12:06

Actually, Luboš, if you don't mind I have a quick follow-up question: is this set of functions completely spherically symmetric? It's just that I have seen mention that the "helium nucleus is spherically symmetric", and I am now wondering whether this statement is as much an approximation as the proton structure diagram.
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romkynsJan 14 '11 at 12:11

@romkyns: that picture is indeed misleading if interpreted literally. But it captures few essential points: quark content of the proton (determining e.g. the total charge of proton) and the interaction modeled by springs. Springs are indeed the best way to imagine strong interaction at low energies. As you pull the quarks apart, the energy increases and they want to go back together. The more you pull them, the stronger the force. This is in contrast with Coulomb force which gets weaker at long distances (which is one of the reasons why you can model an atom with just quantum mechanics).
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MarekJan 15 '11 at 12:44

Dear Romkyns, if I had to draw a scheme of the proton on paper, without equations, I would probably draw the same thing haha. Again, the proton's wave functions - even as a function of the non-relativistic quarks in an approximation - has many components given by the quarks' spins. They are different and their shape in space also depends on the 3 spins. Finally, the whole proton - including all the components of the wave functions in space - also fails to be spherically symmetric. After all, it has a nonzero spin so it can't be symmetric under all rotations. It also has a mag. dipole moment.
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Luboš MotlJan 17 '11 at 20:13

And I propose to look at my own pictures here. There is no proton wave function in atom but a wave function of the relative motion. It is a quasi-particle wave function. Proton is in a mixed state when in atom.

Concerning always bound quarks, it is a non-linear problem with a strong coupling.

And apart from the wave function with many arguments, one can think of the "charge density" $ \rho$(r) determined with the charge form-factor(s) F(q). This is more tangible.

At a very naive level your question amounts to asking what the bound state for three bodies should look like. I would suggest looking up "Efimov trimers". These are many-body bound states predicted by Efimov some 40 years ago and recently experimentally observed. I'm not suggesting that their is a one-to-one correspondence between Efimov states and bound states of nucleons, but they should at the very least provide some sort of a picture.